Question

At equilibrium, both the forward reaction and the reverse reaction occur simultaneously and at the same rate. Since the rates are the same, the concentrations of the reactants and products do not change. An equilibrium constant expression can be developed, which is a mathematical expression indicating the relationship between the reactants and the products. The equilibrium concentrations can be substituted into the equilibrium expression to determine the equilibrium constant, K. The equilibrium constant can also be denoted as "K", "Kc" or "Keq". What is the concentration of NO, if the concentration of N2 and O2 were measured to be 152 M, and the equilibrium constant, K, is 2.0 × 10-9?

Answer 1

Answer : The concentration of NO is, $6.8\times 10^{-3}M$

Solution :  Given,

Concentration of $N_2$ and $O_2$ = 152 M

Equilibrium constant, $K_c$ = $2.0\times 10^{-9}$

The given equilibrium reaction is,

$N_2(g)+O_2(g)\rightleftharpoons 2NO(g)$

The expression of $K_c$ will be,

$K_c=\frac{[NO]^2}{[N_2][O_2]}$

Now put all the given values in this expression, we get:

$2.0\times 10^{-9}=\frac{[NO]^2}{152\times 152}$

$[NO]=6.8\times 10^{-3}M$

Therefore, the concentration of NO is, $6.8\times 10^{-3}M$